1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
use std::fmt;
use std::iter::FusedIterator;

use super::lazy_buffer::LazyBuffer;
use alloc::vec::Vec;

use crate::adaptors::checked_binomial;

/// An iterator to iterate through all the `k`-length combinations in an iterator.
///
/// See [`.combinations()`](crate::Itertools::combinations) for more information.
#[must_use = "iterator adaptors are lazy and do nothing unless consumed"]
pub struct Combinations<I: Iterator> {
    indices: Vec<usize>,
    pool: LazyBuffer<I>,
    first: bool,
}

impl<I> Clone for Combinations<I>
where
    I: Clone + Iterator,
    I::Item: Clone,
{
    clone_fields!(indices, pool, first);
}

impl<I> fmt::Debug for Combinations<I>
where
    I: Iterator + fmt::Debug,
    I::Item: fmt::Debug,
{
    debug_fmt_fields!(Combinations, indices, pool, first);
}

/// Create a new `Combinations` from a clonable iterator.
pub fn combinations<I>(iter: I, k: usize) -> Combinations<I>
where
    I: Iterator,
{
    Combinations {
        indices: (0..k).collect(),
        pool: LazyBuffer::new(iter),
        first: true,
    }
}

impl<I: Iterator> Combinations<I> {
    /// Returns the length of a combination produced by this iterator.
    #[inline]
    pub fn k(&self) -> usize {
        self.indices.len()
    }

    /// Returns the (current) length of the pool from which combination elements are
    /// selected. This value can change between invocations of [`next`](Combinations::next).
    #[inline]
    pub fn n(&self) -> usize {
        self.pool.len()
    }

    /// Returns a reference to the source pool.
    #[inline]
    pub(crate) fn src(&self) -> &LazyBuffer<I> {
        &self.pool
    }

    /// Resets this `Combinations` back to an initial state for combinations of length
    /// `k` over the same pool data source. If `k` is larger than the current length
    /// of the data pool an attempt is made to prefill the pool so that it holds `k`
    /// elements.
    pub(crate) fn reset(&mut self, k: usize) {
        self.first = true;

        if k < self.indices.len() {
            self.indices.truncate(k);
            for i in 0..k {
                self.indices[i] = i;
            }
        } else {
            for i in 0..self.indices.len() {
                self.indices[i] = i;
            }
            self.indices.extend(self.indices.len()..k);
            self.pool.prefill(k);
        }
    }

    pub(crate) fn n_and_count(self) -> (usize, usize) {
        let Self {
            indices,
            pool,
            first,
        } = self;
        let n = pool.count();
        (n, remaining_for(n, first, &indices).unwrap())
    }
}

impl<I> Iterator for Combinations<I>
where
    I: Iterator,
    I::Item: Clone,
{
    type Item = Vec<I::Item>;
    fn next(&mut self) -> Option<Self::Item> {
        if self.first {
            self.pool.prefill(self.k());
            if self.k() > self.n() {
                return None;
            }
            self.first = false;
        } else if self.indices.is_empty() {
            return None;
        } else {
            // Scan from the end, looking for an index to increment
            let mut i: usize = self.indices.len() - 1;

            // Check if we need to consume more from the iterator
            if self.indices[i] == self.pool.len() - 1 {
                self.pool.get_next(); // may change pool size
            }

            while self.indices[i] == i + self.pool.len() - self.indices.len() {
                if i > 0 {
                    i -= 1;
                } else {
                    // Reached the last combination
                    return None;
                }
            }

            // Increment index, and reset the ones to its right
            self.indices[i] += 1;
            for j in i + 1..self.indices.len() {
                self.indices[j] = self.indices[j - 1] + 1;
            }
        }

        // Create result vector based on the indices
        Some(self.indices.iter().map(|i| self.pool[*i].clone()).collect())
    }

    fn size_hint(&self) -> (usize, Option<usize>) {
        let (mut low, mut upp) = self.pool.size_hint();
        low = remaining_for(low, self.first, &self.indices).unwrap_or(usize::MAX);
        upp = upp.and_then(|upp| remaining_for(upp, self.first, &self.indices));
        (low, upp)
    }

    #[inline]
    fn count(self) -> usize {
        self.n_and_count().1
    }
}

impl<I> FusedIterator for Combinations<I>
where
    I: Iterator,
    I::Item: Clone,
{
}

/// For a given size `n`, return the count of remaining combinations or None if it would overflow.
fn remaining_for(n: usize, first: bool, indices: &[usize]) -> Option<usize> {
    let k = indices.len();
    if n < k {
        Some(0)
    } else if first {
        checked_binomial(n, k)
    } else {
        // https://en.wikipedia.org/wiki/Combinatorial_number_system
        // http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf

        // The combinations generated after the current one can be counted by counting as follows:
        // - The subsequent combinations that differ in indices[0]:
        //   If subsequent combinations differ in indices[0], then their value for indices[0]
        //   must be at least 1 greater than the current indices[0].
        //   As indices is strictly monotonically sorted, this means we can effectively choose k values
        //   from (n - 1 - indices[0]), leading to binomial(n - 1 - indices[0], k) possibilities.
        // - The subsequent combinations with same indices[0], but differing indices[1]:
        //   Here we can choose k - 1 values from (n - 1 - indices[1]) values,
        //   leading to binomial(n - 1 - indices[1], k - 1) possibilities.
        // - (...)
        // - The subsequent combinations with same indices[0..=i], but differing indices[i]:
        //   Here we can choose k - i values from (n - 1 - indices[i]) values: binomial(n - 1 - indices[i], k - i).
        //   Since subsequent combinations can in any index, we must sum up the aforementioned binomial coefficients.

        // Below, `n0` resembles indices[i].
        indices.iter().enumerate().try_fold(0usize, |sum, (i, n0)| {
            sum.checked_add(checked_binomial(n - 1 - *n0, k - i)?)
        })
    }
}